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The Hamiltonian is given by the sum over links $l_i$

$$\mathcal{H} = \sum_i l_i \epsilon$$

where $l_i = 0,1$. Hence the partition function is

$$Z_N = \sum_{l_1 = 0,1} \sum_{l_2 = 0,1} \cdots \sum_{l_N = 0,1} e^{-\beta \epsilon l_1} e^{-\beta \epsilon l_2} \cdots  e^{-\beta \epsilon l_N}$$

In the case of independent links, this readily factorises into

$$Z_N = \sum_{l_1 = 0,1} e^{-\beta \epsilon l_1} \sum_{l_2 = 0,1} e^{-\beta \epsilon l_2} \cdots \sum_{l_N = 0,1} e^{-\beta \epsilon l_N} = \left[ \sum_{l = 0,1} e^{-\beta \epsilon l} \right]^N = (1+e^{-\beta \epsilon})^N$$

The number of open links is proportional to the total energy $E = n\epsilon$, hence if we calculate the average energy

$$\langle E \rangle = - \frac{\partial}{\partial \beta} \log Z_N = - N ( 1 + e^{-\beta \epsilon})^{-1} \frac{\partial}{\partial \beta} e^{-\beta \epsilon} = N \epsilon (e^{\beta \epsilon} + 1)^{-1}$$

then immediately

$$\langle n \rangle = \frac{1}{\epsilon} \langle E \rangle = N (1 + e^{\beta \epsilon})^{-1}$$

Thus at low temperatures $\beta \rightarrow \infty$ and

$$\langle n \rangle_{T \rightarrow 0} = 0$$

and at high temperatures $\beta \rightarrow 0$ thus

$$\langle n \rangle_{T \rightarrow \infty} = N / 2$$

In equilibrium, the free energy

$$F = E - TS$$

of a system is minimised. Hence at low $T$, energy tends to be minimised, and at high $T$, entropy tends to be maximised. In the case of the independent zipper we see that at low temperatures energy is indeed minimised $\langle E \rangle = \langle n \rangle \epsilon = 0$. For the indepent zipper the entropy proportional to the number of ways to choose $n$ open links from $N$ links, which is maximised when $n = N/2$, hence the entropy of the independent zipper is indeed maximised at high $T$.

Now if the link is allowed to open only from one end, the partition function becomes

$$Z_N = \sum_{n=0}^N e^{-\beta \epsilon n} = \frac{ 1 - e^{ - \beta \epsilon (N+1)} }{ 1 - e^{- \beta \epsilon} }$$

where we have used

$$ \sum_{n=0}^N x^{an} = \frac{ 1 - x^{a(N+1)} }{ 1 - x^a }$$

and similarly we can find the expected $n$ by considering the expected $E$

$$\langle E \rangle = - \frac{ \partial }{ \partial \beta } \log Z_N = \epsilon \left[ \frac{ e^{-\beta \epsilon }{ 1 - e^{- \beta \epsilon} } - \frac{(N+1)e^{- \beta \epsilon(N+1)}}{ 1 - e^{- \beta \epsilon (N+1) } } \right]$$

hence rather trivially we obtain

$$\langle n \rangle = \left[ \frac{ e^{-\beta \epsilon }{ 1 - e^{- \beta \epsilon} } - \frac{(N+1)e^{- \beta \epsilon(N+1)}}{ 1 - e^{- \beta \epsilon (N+1) } } \right]$$

In the low-temperature limit $T \rightarrow 0 \Rightarrow \beta \rightarrow \infty$ and hence $e^-\beta \epsilon \rightarrow 0$ and we recover

$$\langle n \rangle_{T \rightarrow 0} = 0$$

In the high temperature limit $T \rightarrow \infty \Rightarrow \beta \rightarrow 0$, but we must carefully expand the exponentials to find the limit

$$\langle n \rangle_{T \rightarrow 0} = \frac{1 - \beta \epsilon}{\beta \epsilon} - \frac{ (N+1)(1 - \beta \epsilon(N+1)) }{ \beta \epsilon (N+1) }$$

where we have discarded terms $\mathcal{O}(\beta^2)$ which will vanish in the desired limit, hence

$$\langle n \rangle_{T \rightarrow \infty} = N$$

Intepreting this result in terms of energy-entropy competition, in the low-temperature limit the zipper is completely closed as before, minimising the energy. However he entropy of the restricted zipper does not depend upon $n$,

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